In a textbook on fractional calculus, the following equality is stated without any comments or a proof:
$$ \frac{d}{dt} \left( \int\limits_{-\infty}^{t} (t-s)^{-\alpha} f(s) ds \right) = \alpha \int\limits_{0}^{\infty} \frac{f(t)-f(t-s)}{s^{\alpha + 1}}ds $$
where $0 <\alpha < 1$, $t \in \mathbb{R}$ and $f$ is absolutely continuous. It is also assumed that the LHS integral exists. Could anyone provide any hints as to how to get from the LHS integral to the RHS integral? I've tried using a substitution $t-s = z$, singling out the singularity at $s = t$, but I couldn't get to the RHS. Any hints would be highly appreciated.
let: $$\mathcal{F}(t,s)=(t-s)^{-\alpha}f(s)$$
by the liebniz integral rule: $$\frac{\mathrm{d}}{\mathrm{d}t}\int_{-\infty}^t\mathcal{F}(t,s)\,\mathrm{d}s=\mathcal{F}(t,t)+\int_{-\infty}^t\frac{\partial\mathcal F(t,s)}{\partial t}\mathrm{d}s$$
and:
$$\frac{\partial\mathcal F}{\partial t}=-\alpha(t-s)^{-(\alpha+1)}f(s)$$
From the existence of the original integral, it is implied that equally $\mathcal{F}(t,t)$ exists. I believe that if you can define this in terms of the integral and make a new substitution of $s\mapsto t-s$ you can reach the RHS, I will continue working on it